How many nonabelian groups of order 2009? (Check work)

Question

I just need someone to check this argument.

Let $G$ be a nonabelian group of order $2009$. The prime factorization of $2009$ is $7^2 \cdot 41$. Let $n$ be the number of Sylow 7-subgroups.

Then $n \equiv 1$ mod $7$ and $n$ divides $41$. Since $41$ is prime, we must have $n =1$ or $n=41$, but $41 \equiv 6$ mod $7$. Thus $n=1$ and we have a unique Sylow 7-subgroup $H$.

Since $H$ is the unique Sylow 7-subgroup, we have that $H$ is a normal subgroup of $G$. Taking the quotient we have that $|G/H|=41$, so $G/H$ is cyclic of order 41.

Since $G/H$ is cyclic, $G$ is abelian. Thus there is no nonabelian group of order $2009$.

Answer 1

If $G/H$ is cyclic then you can not say $G$ is abelian as Mikko pointed out.

You should go on like that;

Let $q=41$ then $n_q$ must divides $49$ and $n_q\equiv 1 \ mod \ (41) \implies n_q=1 $.

Thus all sylow subgroup is normal, $G\cong H\times K$ where $|H|=49$ and $|K|=41$ since any group of order $p^2$ is abelian, $H$ is abelian then clearly $G$ is abelian.

Answer 2

I posted an erroneous remark in a (in the meantime deleted) answer - (thanks Mikko for pointing it out). Nevertheless, the arguments given by Mesel and Frost Boss nicely generalize to groups $G$ of order $p^2q$, with $p$, and $q$ primes, such that $p \lt q$, $p \nmid q-1$ and $q \nmid p^2-1$. Such a group has to be abelian.
(Note the conditions $p \lt q$, $p \nmid q-1$ imply $q \nmid p^2-1$, so $p \lt q$, $p \nmid q-1$ is sufficient to guarantee the abelianess).